Exercise 8.1 · Q3 (MCQ): First-quadrant area of $x^2 + y^2 = 4$ between $x = 0$ and $x = 2$

Basics you need

Solution

1. Identify the region and sketch it

   $x^2 + y^2 = 4 = 2^2$ is the circle with centre $O(0, 0)$ and radius 2. It meets the positive axes at $(2, 0)$ and $(0, 2)$⁠.

   In the first quadrant ($x \ge 0$⁠, $y \ge 0$⁠), the region is bounded by the arc of the circle, the line $x = 0$ (the $y$-axis) and the line $x = 2$⁠, which touches the circle at $(2, 0)$⁠. Its lower edge is the $x$-axis, the edge of the first quadrant. So the region is the quarter disc $OAB$ with $A(2, 0)$⁠, $B(0, 2)$⁠.

   The line $x = 2$ meets the circle only at $(2, 0)$⁠, so in the first quadrant the region reaches from $x = 0$ to $x = 2$ under the arc: the whole first-quadrant quarter of the disc.
   The circle $x^2 + y^2 = r^2$ has centre $O$ and radius $r$⁠. Lesson: circles and ellipses

2. Solve for $y$

   $y^2 = 4 - x^2$⁠, so $y = \pm\sqrt{4 - x^2}$⁠. In the first quadrant $y \ge 0$⁠, so the top edge is $y = \sqrt{4 - x^2}$⁠, $0 \le x \le 2$⁠.

   The top of each vertical strip is on the upper semicircle, where $y$ is the positive root.
   Take the positive root above the $x$-axis. Basics: solve for y to get the top edge

3. Write the area as an integral

   $$\text{Area} = \int_0^2 y\,dx = \int_0^2\sqrt{4 - x^2}\,dx.$$
   Vertical strips of height $y = \sqrt{4 - x^2}$ and width $dx$⁠, from the line $x = 0$ to the line $x = 2$⁠. The integrand is $\ge 0$⁠, so no part is below the axis.
   Section 8.2: $A = \int_a^b y\,dx$⁠. Lesson: the area under a curve

4. Integrate with the square-root formula

   With $a = 2$⁠, $\frac{a^2}{2} = 2$⁠:

   $$\begin{aligned} \text{Area} &= \Big[\frac{x}{2}\sqrt{4 - x^2} \\ &\quad + 2\sin^{-1}\frac{x}{2}\Big]_0^2 \\ &= \big(1 \times 0 + 2\sin^{-1}1\big) \\ &\quad - \big(0 + 2\sin^{-1}0\big) \\ &= 2\cdot\frac{\pi}{2} = \pi. \end{aligned}$$
   At $x = 2$ the root is 0 and $\sin^{-1}1 = \frac{\pi}{2}$ (principal value); at $x = 0$ both terms are 0.
   Section 7.6.2 (iii); Theorem 2, Section 7.8.3. Basics: the quarter-circle integral

5. Check with $\pi r^2$

   The whole disc of radius 2 has area $\pi \cdot 2^2 = 4\pi$ (Example 1), and the region is one quarter of it: $\frac{4\pi}{4} = \pi$⁠. The correct option is (A).

   The circle is symmetrical about both axes, so its four quarters have equal areas. The integral and the formula agree.
   One quarter of the circle's area. Basics: one quarter, times four

6. Examine every option

   A quick bound decides it without any integration: the quarter disc contains the triangle with vertices $O$⁠, $(2, 0)$⁠, $(0, 2)$ (area 2) and lies inside the square of side 2 (area 4). So the area is between 2 and 4.

   Option	Value	Between 2 and 4?	Verdict
   (A) $\pi$	$\approx 3.14159$	Yes	Correct: the integral above
   (B) $\frac{\pi}{2}$	$\approx 1.57080$	No	Half the area
   (C) $\frac{\pi}{3}$	$\approx 1.04720$	No	Too small
   (D) $\frac{\pi}{4}$	$\approx 0.78540$	No	Quarter of a unit circle

   (B) $\frac{\pi}{2}$ is what you get by writing $\frac{a}{2} = 1$ instead of $\frac{a^2}{2} = 2$ in front of $\sin^{-1}$⁠. (D) $\frac{\pi}{4}$ is the quarter of the circle of radius 1: what you get by using the radius 1, or by writing $\frac12\sin^{-1}1$⁠. (C) $\frac{\pi}{3}$ comes from no natural slip; it is simply too small. All three are below 2, the area of the triangle inside the region, so none can be right.
   A region contains everything inside it, so its area is at least the triangle's area. Lesson: why the right option is right

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